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Dipole in a Uniform External Field is in stable equilibrium and is in the direction of the electric field vector and is parallel \(\overrightarrow{P} ||\overrightarrow{E}\). An electric dipole is a pair of electric charges possessing equal magnitude but opposite charges separated by distance.
- An electric dipole is a separation between positive and negative charges of equal magnitudes.
- Dipole in a uniform electric field experiences torque which is a force in rotational motion or an angular force.
- If a dipole is placed in a uniform electric field, the torque will make it rotate.
- The torque aligns the dipole parallel to the direction of the electric field with zero net force.
Read More: Difference between Torque and Moment
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Key Terms: Torque, Dipole, Electric Dipole, Magnetic Dipole, Calculating Torque, Angular frequency, Oscillations, Time period, Electric field, Angular force, Angular velocity
Torque and Its Calculation
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Let us assume that the dipole is introduced to a uniform external field in order to seek out torque experienced by a dipole when placed in an external field. This field will provide an electrical force having qE magnitude in the upward direction for the positive charge, while downward direction for the negative charge.
- We can locate that the dipole rests in translational equilibrium as the net force is zero.
- But what's rotational equilibrium? Taking this case into consideration, it is possible that the dipole might remain in a stationary position but does rotate with a specific angular velocity.
- This fact has been experimentally proven and unveils that both electrostatic forces (qE) work as a torque being applied within the clockwise direction.
- Hence, when placed in the uniform external field, the dipole does get to rotate.
- It is worth noting that torque always operates in a couple.
- Additionally, its magnitude is equal to the resultant product of force and its arm.
- Here, the arm can be considered as the distance between the point at which rotation takes place for the dipole.
Dipole Placed within the Uniform External field
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The origin is considered the point when a dipole is placed in the uniform external field. Moreover, torque is represented by the symbol ‘τ’. It must be remembered that torque is a vector quantity. As per mathematical basis:
Torque magnitude (τ) = q E × d sin θ
⇒ τ = qEd sin θ
But qd = p (dipole moment)
⇒ τ = pE sin θ
Therefore, we will say that the dipole moment and cross product of the electric field refers to the vector form of torque.
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Derivation of Torque of an Electric Dipole
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Derivation of Torque on an Electric Dipole can be understood from the figure given below.
Consider a dipole with charges +q and –q forming a dipole since they are a distance d away from each other. Let it be placed in a uniform electric field of strength E such that the axis of the dipole forms an angle θ with the electric field. Refer to the following image for the derivation.
And the direction of the dipole moment is from the positive to the negative charge; it can be seen from the above equation that the torque is the cross product of the dipole moment and electric field. Notice that the torque is in the clockwise direction (hence negative) in the above figure if the direction of the Electric field is positive.
Oscillation of an Electric Dipole in a Uniform Electric Field
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Consider an electric dipole of dipole moment p placed in uniform electric field E at an angle θ. The torque experienced by the dipole is given by
τ = – pE sinθ
If θ is very small then, sin θ ≈ θ
⇒ τ = – pEθ
Let I will be the moment of inertia of the dipole and α be angular acceleration, then
τ = Iα
From the above two equations, we get
Iα = – pEθ
⇒ α = – (pE/I)θ
Comparing the equation with the equation of simple harmonic motion α = -ω2θ, we get
ω2 = pE/I
Also, ω = 2π/T
⇒ 2π/T = pE/I
\(\Rightarrow \omega = \sqrt {\frac {pE}{I}}\)
But, ω = 2π/T (T is the time period of oscillation of the dipole)
\(\Rightarrow \frac {2\pi} {T} = \sqrt {\frac {pE}{I}}\)
⇒ \( T = 2\pi \sqrt {\frac {I}{pE}}\)
Also, T = 1/f (Where f is the frequency of the oscillation)
⇒ \( f =\frac {1}{T}= \frac {1}{2\pi} \sqrt {\frac {pE}{I}}\)
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Previous Year Questions
- In finding the electric field using Gauss law the formula….[JEE Main 2020]
- two charges of equal amount +Q are placed on a line...[WEBJEE 2016]
- A thin disc of radius b=2a has a concentric hole of….[JEE Main 2015]
- The average velocity and the average speed of the toy car between 0to30to3 seconds are respectively….[NEET 2018]
- The electric flux linked to the surface, in units of volt m is...[NEET 2010]
- maximum torque exerted by the field on the dipole is….[KEAM]
- What happens when some charge is placed on a soap bubble?
- On charging by conduction, mass of a body may…
- Object may acquire an excess or deficiency of charge by….
- In figure, two positive charges, q2 and q3 fixed along the yy axis, exert a net electric force in the +x direction on a charge q1 fixed along the x axis. If a positive charge Q is added at (x, 0), the force on q1
- In a medium of dielectric constant K, the electric field is →E→ . If ∈0 is permittivity of the free space, the electric displacement vector is….
- If an electron has an initial velocity in a direction different from that of an electric field, the path of the electron is…..
- If a body is positively charged, then it has…..
- If a body is charged by rubbing it, its weight…...
- If 109 electrons move out of a body to another body every second, then the time required to get a total charge of 1C on the other body is
- Four point charges are placed at the corners of a square ABCD of side 10cm, as shown in figure. The force on a charge of 1μC placed at the centre of square is….
- Five charges q1, q2, q3, q4 and q5 are fixed at their positions as shown in figure. S is Gaussian surface. The Gauss? law is given by ∮s→E⋅→ds=qε0. Which of the following statements is correct?
- Five styrofoam balls are suspended from insulating threads. Several experiments are performed on the balls and the following observations are made - (i) Ball A repels C and attracts B (ii) Ball D attracts B and has no effect on E (iii) A negatively charged rod attracts both A and E. An electrically neutral styrofoam ball gets attracted if placed nearby a charged body due to induced charge. What are the charges, if any, on each ball?
- Figure below shows a closed surface which intersects a conducting sphere. If a positive charge is placed at the point P, the flux of the electric field through the closed surface…
- Figure shows electric field lines in which an electric dipole →p→ is placed as shown. Which of the following statements is correct?
Things to Remember
- Electric dipole is a pair of electric charges possessing equal magnitude but opposite charges separated by distance.
- In simple words, an electric dipole is a separation between positive and negative charges.
- It is worthy to note that torque always operates in a couple.
- Additionally, its magnitude is equal to the resultant product of force and its arm.
- The torque is the cross product of the dipole moment and electric field.
- The torque is in the clockwise direction (hence negative) in the above figure if the direction of the Electric field is positive.
Sample Questions
Ques. An electric dipole comprises the charges +q and -q separated by a distance of L and is in stable equilibrium in a uniform electric field. The electrostatic potential energy of the dipole is,
(1) qLE
(2) Zero
(3) - qLE
(4) - 2qLE (CBSE 2020)
Ans. The electrostatic potential energy of the dipole is - qLE.
Explanation:-
Given, an electric dipole consists of the charges +q and -q,
The separation between these two charges is L,
As we know the electrostatic potential energy of the dipole that is,
P = -pE … (1)
Here, p = electric dipole which is defined by the product of charge and distance between them.
p = qL
By putting the value of p in equation 1 we get,
P = -qLEP = -qLE
Thus, this is the required solution.
Ques. Give reasons for why there must be an electrostatic field at the surface of a charged conductor be normal to the surface at every point? (Foreign 2014, Delhi 2012)
Ans. The work done in moving a charge from one point to another is always zero on an equipotential surface. The electric field will have a non-zero component along the surface which would cause work to be done in moving a charge on an equipotential surface if electric field is not normal.
Ques. (a) Draw equipotential surfaces that are corresponding to a uniform electric field in the z-direction.
(b) Obtain an expression for the electric potential at any point along the axial line of an electric dipole. (CBSE 2019)
Ans. (a) As we know that the electric field strength is proportional to the distance between the surface and so the surface is evenly paced.
In a way that the electric field is always perpendicular to the equipotential surface. Therefore, the equipotential surfaces will be infinite plane sheets in the z direction parallel to the x-y plane.
(b) Let us assume that an electric dipole having -q and q charges separated by a certain distance 2a.
Now, the electric potential at any point along an axial line of an electric dipole:
Let the distance of the point P along the axial line from the centre of the dipole is R
Thus, the electric potential at P due to charge A will be V1 and the electric potential at point P due to charge b will be V2
Calculating the net potential at point P,
We get,
We know the dipole moment is
Ques. What is electric flux and write its S.I unit. The electric field components in the figure shown are, , Assuming, , calculate the charge within the cube. (CBSE 2018)
Ans. The electric flux through an area can be defined as the number of field lines passing normally through an area
Electric flux,
Ques. (i) Obtain the expression for the torque τ experienced by an electric dipole of dipole moment p in a uniform electric field, E.
(ii) What would happen if the field were not uniform? (CBSE Delhi 2017)
Ans: (i) Consider a dipole with charges +q and –q forming a dipole since they are a distance d away from each other. Let it be placed in a uniform electric field of strength E such that the axis of the dipole forms an angle θ with the electric field. Refer to the following image for the derivation.
(ii) If the electric field is not uniform, the net force acting on that dipole will not be zero. In such a case the above derivation will not hold true.
Ques. (i) Obtain the expression of an electric field at a point on the equatorial line of an electric dipole.
(ii) Depict the orientation of the dipole in (a) stable and (b) unstable equilibrium in a uniform electric field. (Delhi 2017)
Ans. (i) In an electric dipole moment, it is the product of the magnitude of either charge or distance between them.
The direction of the dipole moment is from negative to positive charge.
Due to +q charge, electric field intensity at P is
Due to -q charge, electric field intensity at P is
(ii) (a) For stable equilibrium, the angle between p and E is 0 degree
Ques. Give a real-life example of a dipole and electric field. (3 Marks)
Ans. Try to comb dry hair and quickly bring the comb close to the paper pieces. The comb will pull the paper pieces because it acquires charge due to induction. On the contrary, the comb can polarize the paper pieces, i.e., produce a net dipole moment (direction of the electric field). Since the electric field stays non-uniform, the pieces of paper get attracted in the direction of the comb.
Ques. How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased? (2 Marks) [Delhi 2016]
Ans. The electric flux due to a point charge enclosed by a spherical Gaussian surface remains ‘unaffected’ when its radius is increased.
Ques. Why are electric field lines perpendicular at a point on an equipotential surface of a conductor? (3 Marks) [Comptt. All India 2015]
Ans. If the electric field lines were not normal to the equipotential surface, they would have a non-zero component along the surface. To move a unit test charge against the direction of the component of the field, work would have to be done which means this surface cannot be an equipotential surface. Thus, electric field lines are perpendicular at a point on an equipotential surface of a conductor.
Ques. Why do the electrostatic field lines not form closed loops? (2 Marks) [All India 2014]
Ans. Electric field lines do not form closed loops because the direction of an electric field is from positive to negative charge. So one can regard a line of force starting from a positive charge and ending on a negative charge. This indicates that electric field lines do not form closed loops.
Ques. What is the direction of the electric field at the surface of a charged conductor having charge density σ < 0? (2 Marks) [Comptt. Delhi 2012]
Ans. The direction of the electric field is normal and inward to the surface.
Ques. Why must the electrostatic field be normal to the surface at every point of a charged conductor? (2 Marks) [Delhi 2012]
Ans. The electrostatic field should be normal to the surface so that tangent on charged conductor gives the direction of the electric field at that point.
Ques. Name the physical quantity whose S.I. unit is JC-1. Is it a scalar or a vector quantity? (1 Mark) [All India 2010]
Ans. A physical quantity whose S.I. unit is JC-1 is Electric potential. It is a Scalar quantity.
Ques. (a) By using Gauss Law, derive an expression for an electric field due to a spherical shell of uniform charge distribution and radius R at a point lying at a distance x from the center of the shell, such that
(i)
(ii)
(b) An electric field is uniform and acts along the positive x direction for the region positive x. It is also uniform with the same magnitude but acts in a negative x-direction for negative x. The value of the field is E = 200N/C for x>0 and E= - 200N/C for x<0. A right circular cylinder of length 20 cm and radius 5 cm has its center at the origin and its axis along the x-axis so that one flat face is at x = +10 cm and the other is at x= -10 cm.
Find, (i) The net outward flux through the cylinder.
(ii) The net charge present inside the cylinder. (CBSE 2020)
Ans. (a) (i)
Inside a spherical shell, the charge is zero, and therefore the electric field E = 0
(ii) x > R
Let us consider a point P outside the shell at a distance r.
Now, let us assume a spherical Gaussian surface of radius r.
Then by Gauss law,
Flux enclosed by the surface
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